Core Connections Algebra 2, 2013
CC
Core Connections Algebra 2, 2013 View details
3. Section 8.3
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Exercise 121 Page 411

Use an area model to divide the polynomial.

x_1=1
x_2=- 3
x_3=0.5

Practice makes perfect
Since the graph crosses the x-axis at x=1, we know that (x-1) is a factor. (x-1) (other factor)=2x^3+3x^2-8x+3 Let's use an area model to find the other factor. The first term of the other factor must be 2x^2. It must be this in order for the product of our factor's first terms to equal 2x^3. ( x-1) ( 2x^2......)= 2x^3+3x^2-8x+3

With this information, we can begin creating our area model's first column.

In the original equation we have the term 3x^2. Since one tile of our area model contains - 2x^2, we must add 5x^2 to get a sum of 3x^2. With this information, we can identify the second term of our factor and the contents of the area model's second column.

Again, examining the original equation we find the term - 8x. Since one tile of our area model contains - 5x, we must add -3x to get a sum of - 8x. With this information, we can identify the third term of our factor and the contents of the third column.

If we add all of the terms contained within the area model, the sum should equal the expression on the left-hand side of the original equation. 2x^3+(- 2x^2)+5x^2+(- 5x)+(- 3x)+3 ⇓ 2x^3+3x^2-8x+3 Now we know that the other factor is (2x^2+5x-3). With this information, we can rewrite the original equation. (x-1) (2x^2+5x-3)=0 We can find the remaining roots by solving for x in the second factor. Let's use the Quadratic Formula.
2x^2+5x-3=0
x=- 5 ± sqrt(5^2-4 * 2( - 3))/2 * 2
Solve for x
x=- 5 ± sqrt(25+24)/4
x=- 5 ± sqrt(49)/4
x=-5 ± 7/4
lx=.-12 /4. x=.2 /4.
lx_1=- 3 x_2=0.5
The two other roots are x=- 3 and x=0.5, which means their corresponding factors are (x+3) and (x-0.5). Now we can fully factor our expression. Let's also summarize all of the zeroes. (x-1)(x+3)(x-0.5)=0 [1em] Zeroes: x_1=1, x_2=- 3, x_3=0.5